14.Probability
medium

A box contains $15$ tickets numbered $1, 2, ....... 15$. Seven tickets are drawn at random one after the other with replacement. The probability that the greatest number on a drawn ticket is $9$, is

A

${\left( {\frac{9}{{10}}} \right)^6}$

B

${\left( {\frac{8}{{15}}} \right)^7}$

C

${\left( {\frac{3}{5}} \right)^7}$

D

None of these

(IIT-1983)

Solution

(c) On trial, $n = 15$ since any of the $15$ numbers can be on the selected coin and $m = 9$ since the largest number is $9$ and so it can be $1$ or $2$ or $3………$ or $9$.
We have required probability $ = {\left( {\frac{9}{{15}}} \right)^7} = {\left( {\frac{3}{5}} \right)^7}.$

Standard 11
Mathematics

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